From the circuit diagram, it can be said that the diode is reverse biassed, with applied voltage of 6.0V. Under reverse bias condition, Current in the circuit = Drift current So, the current in the circuit is 20µA. Voltage across the diode will be equal to the voltage of the battery minus the voltage drop across the 20 ohm resistor. ⇒V=6−iR ⇒V=6−(20×20×10−6) ⇒V=6−(4×10−4) ⇒V=10−4(60000−4) ⇒V=59996×10−4 ⇒V=5.9996V≅5V